Question: Complete the square to solve for $x$. $x^{2}+7x-8 = 0$
Solution: Move the constant term to the right side of the equation. $x^2 + 7x = 8$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $7$ , so half of it would be $\dfrac{7}{2}$ , and squaring it gives us ${\dfrac{49}{4}}$ $x^2 + 7x { + \dfrac{49}{4}} = 8 { + \dfrac{49}{4}}$ We can now rewrite the left side of the equation as a squared term. $( x + \dfrac{7}{2} )^2 = \dfrac{81}{4}$ Take the square root of both sides. $x + \dfrac{7}{2} = \pm\dfrac{9}{2}$ Isolate $x$ to find the solution(s). $x = -\dfrac{7}{2}\pm\dfrac{9}{2}$ The solutions are: $x = 1 \text{ or } x = -8$ We already found the completed square: $( x + \dfrac{7}{2} )^2 = \dfrac{81}{4}$